Question

solve this question otherwise I will report you​

Answer 1

Questions

(i) Evaluate ⇒ $\frac{10}{3}-5\frac{1}{5}+3\frac{5}{6}$

(ii) Evaluate ⇒ $(\frac{3}{4}\times \frac{2}{5}) + (\frac{-9}{11}\times \frac{3}{4}) - (\frac{3}{4}\times \frac{-1}{5})$

(iii) Find the value of [(-5)⁻²]⁻¹

(iv) Simplify ⇒ $(\frac{5}{4})^{2}\div [(\frac{1}{3})^{-2} \div (\frac{1}{8})^{-1}]^{-1}$

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Answers

(i) We will apply BODMAS here. We will do addition first and then subtraction.

⇒ $\frac{10}{3}-5\frac{1}{5}+3\frac{5}{6}$

⇒ $\frac{10}{3}+3\frac{5}{6}-5\frac{1}{5}$

Step 1: Convert the mixed fraction to improper fraction. To obtain the numerator of the improper fraction we must multiply the whole number part to the denominator of the fractional part, then we shall add the product to the numerator of the fractional part. For the denominator of the improper fraction it would be the same as the denominator of the fractional part in the mixed fraction.

⇒ $\frac{10}{3}+3\frac{5}{6}-5\frac{1}{5}$

⇒ $\frac{10}{3}+\frac{3\times 6+5 }{6}-\frac{5\times 5+1 }{5}$

⇒ $\frac{10}{3}+\frac{23}{6}-\frac{26 }{5}$

Step 2: Find the LCM of the denominators and make them equal and solve.

$\begin{array}{r | l} 2&3,6,5 \\ \cline{2-2} 3 &3,3,5 \\ \cline{2-2} 5 &1,1,5 \\ \cline{2-2} &1,1,1 \\ \end{array}$

LCM of the denominators ⇒ 2×3×5 = 30

⇒ $\frac{10\times 10 }{3\times 10 }+\frac{23\times 5 }{6\times 5}-\frac{26\times 6 }{5\times 6}$

⇒ $\frac{100 }{30 }+\frac{115 }{30}-\frac{156 }{30}$

⇒ $\frac{215 }{30 }-\frac{156 }{30}$

⇒ $\frac{59 }{30 }$

⇒ $1\frac{29}{30}$

$\bf \therefore \frac{10}{3}-5\frac{1}{5}+3\frac{5}{6} = 1\frac{29}{30}$

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(ii) We will apply BODMAS here. First we will do the brackets and then addition and at last subtraction.

⇒ $(\frac{3}{4}\times \frac{2}{5}) + (\frac{-9}{11}\times \frac{3}{4}) - (\frac{3}{4}\times \frac{-1}{5})$

⇒ $(\frac{3}{4\div 2}\times \frac{2\div 2}{5}) + (\frac{-9}{11}\times \frac{3}{4}) - (\frac{3}{4}\times \frac{-1}{5})$

⇒ $(\frac{3}{2}\times \frac{1}{5}) + (\frac{-9}{11}\times \frac{3}{4}) - (\frac{3}{4}\times \frac{-1}{5})$

⇒ $\frac{3}{10}+ \frac{-27}{44}- \frac{-3}{20}$

Step 1: Find LCM of the denominators.

$\begin{array}{r | l} 2& 10,44,20\\ \cline{2-2} 2 & 5,22,10 \\ \cline{2-2} 5& 5,11,5\\ \cline{2-2}11 &1,11,1 \\ \cline{2-2} & 1,1,1\\ \end{array}$

LCM of the denominators ⇒ 2×2×5×11 = 220

Step 2: Make the denominators equal and solve.

⇒ $\frac{3 \times 22}{10 \times 22 }+ \frac{-27 \times 5}{44 \times 5}- \frac{-3 \times 11 }{20 \times 11}$

⇒ $\frac{66}{220 }+ \frac{-135}{220}- \frac{-33 }{220}$

⇒ $\frac{-69}{220 }- \frac{-33 }{220}$

⇒ $\frac{-36}{220}$

⇒ $\frac{-9}{55}$

$\bf \therefore (\frac{3}{4}\times \frac{2}{5}) + (\frac{-9}{11}\times \frac{3}{4}) - (\frac{3}{4}\times \frac{-1}{5}) = \frac{-9}{55}$

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(iii) For this we need to apply laws of exponents.

Step 1: Apply this law of exponent first ⇒ $(a^{m})^{n} = a^{m\times n}$

⇒ [(-5)⁻²]⁻¹

⇒ (-5)⁻²⁽⁻¹⁾

Step 2: Apply this law of integers for exponents ⇒ (-)(-) = (+)

⇒ (-5)⁻²⁽⁻¹⁾

⇒ (-5)²

⇒ (-5) × (-5)

⇒ 25

∴ [(-5)⁻²]⁻¹ = 25

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(iv) For this question we must apply laws of exponents.

Step 1: Apply this law of exponent ⇒ $a^{-m} = \frac{1}{a^{m}}$

⇒ $(\frac{5}{4})^{2}\div [(\frac{1}{3})^{-2} \div (\frac{1}{8})^{-1}]^{-1}$

⇒ $(\frac{5}{4})^{2}\div [3^{2} \div 8^{1}]^{-1}$

⇒ $(\frac{5}{4})^{2}\div [3^{2} \times\frac{1}{8} ]^{-1}$

⇒ $(\frac{5}{4})^{2}\div [\frac{3^{2}}{8} ]^{-1}$

⇒ $(\frac{5}{4})^{2}\div \frac{8}{3^{2}}$

Step 2: Apply this law of exponent ⇒ $(\frac{a}{b})^{m} = \frac{a^{m}}{b^{m}}$

⇒ $(\frac{5}{4})^{2}\div \frac{8}{3^{2}}$

⇒ $\frac{5^{2}}{4^{2}}\div \frac{8}{3^{2}}$

⇒ $\frac{5^{2}}{2^{4}}\div \frac{2^{3}}{3^{2}}$

⇒ $\frac{5^{2}}{2^{4}}\times \frac{3^{2}}{2^{3}}$

Step 3: Apply this law of exponent ⇒ $a^{m}\times a^{n}=a^{m+n}$

⇒ $\frac{5^{2}}{2^{4}}\times \frac{3^{2}}{2^{3}}$

⇒ $\frac{5^{2}}{2^{4+3}}\times 3^{2}$

⇒ $\frac{5^{2}}{2^{7}}\times 3^{2}$

Step 4: Apply this law of exponent ⇒ $a^{m}\times b^{m} = (ab)^{m}$

⇒ $\frac{5^{2}}{2^{7}}\times 3^{2}$

⇒ $\frac{(5\times 3) ^{2}}{2^{7}}$

⇒ $\frac{15^{2}}{2^{7}}$

⇒ $\frac{225}{128}$

$\bf \therefore (\frac{5}{4})^{2}\div [(\frac{1}{3})^{-2} \div (\frac{1}{8})^{-1}]^{-1} = \frac{225}{128}$

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